# Chemistry Question

I was wondering if anyone could point me in the right direction for this question ?

Mg + H2SO4 ----> MgSO4 + H2

Cacluate the minimum mass of magnesium metal that would be required to completely neutralise 0.50 L of 1.0 mol l-1 sulfuric acid.

My calculations so far ...

Mr(H2SO4) = 2xAr(H) + Ar(S) + 4xAr(O)
= 98

Mr(MG) = 24.31

Molarity = 1.0mol / 0.50L = 2

But I am at a loss as to how I get to the final answer .....
Any help would be most appreciated !

## Answers and Replies

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siddharth
Homework Helper
Gold Member
From the reaction you have written, 1 mole of H2S04 is neutralised by 1 mol of Mg. So, what is the number of moles of H2S04 in .5 L? And this will require how many moles of Mg?

Thanks for the reply,

It states that there is 1.0mol of sulfuric acid so the amount of magnesium required would be 1.0mol as well ?

siddharth
Homework Helper
Gold Member
titchwatt said:
Thanks for the reply,

It states that there is 1.0mol of sulfuric acid so the amount of magnesium required would be 1.0mol as well ?
No, it states that there is 1.0 mol per litre of sulphuric acid. So, in 0.5 litres, how many moles are there? That will be equal to the number of moles of Mg required.

Once you find the number of moles of Mg required, do you know how to find the mass of Mg? (Hint: What is the relation between atomic weight, number of moles and mass of the substance?)

That has helped a lot ! Thankyou.